Since nobody else seems to be working on this, let me start by posting a translation of your earlier encoded answer:
There are five hundred twelve patterns. Each row and column is either flipped or not giving two to the tenth patterns. However, these patterns do not all have distinct color patterns. In particular, flipping every row and column is the same as not flipping any of them. There are thus two flipping patterns for every color pattern, giving two to the ninth color patterns. To get the equivalent flipping pattern for a given one, just add the pattern of flipping all rows and columns, resulting in a kind of negative of the original. For example, the equivalent of flipping the first three rows and no columns is flipping the last two rows and all the columns.
My objection to all of this had been that the sequence of moves matters, i.e., that flipping any row followed by any column is non-transitive—reversing the moves matters. This, however, appears not to be the case—or at least I haven’t found a counter-example. If I could prove that all such operations are pairwise transitive, despite the counter-intuitive feeling I have about it, then I accept your solution.
Is there some key bit of insight leading to an obvious mathematical proof? This is starting to bug me! Interesting problem, btw.